Approximaths

Regularization III

In this post, we justify the regularization procedure introduced in previous posts for singular polynomial problems. Consider:

$\displaystyle P(x)=a_0+a_1x+\dots+a_nx^n$

$\displaystyle P(x)=1+A_1x+\dots+A_nx^n$

$\displaystyle P_{\epsilon}(x)=f_0(\epsilon)+f_1(\epsilon)\epsilon^{\alpha_1}A_1x+\dots+f_n(\epsilon)\epsilon^{\alpha_n}A_nx^n$

$\displaystyle P_{\epsilon}(x)=(1+b_0\epsilon+c_0\epsilon^2+\dots)+(1+b_1\epsilon+c_1\epsilon^2+\dots)\epsilon^{\alpha_1}A_1x+\dots+(1+b_n\epsilon+c_n\epsilon^2+\dots)\epsilon^{\alpha_n}A_nx^n$

where $f_0(\epsilon),f_1(\epsilon),\dots,f_n(\epsilon)$ admit power-series expansions in $\epsilon$ , $\alpha_i$ ‘s are rational numbers and $b_i$ ‘s, $c_i$ ‘s are constants.

The last expression is therefore a generalized version of a perturbed polynomial

. We can rearrange this expression as follows:

$\displaystyle P_{\epsilon}(x)=(1+b_0\epsilon+c_0\epsilon^2+\dots)$

$\displaystyle +(1+b_1\epsilon+c_1\epsilon^2+\dots)\epsilon^{\alpha_1}A_1x$

$\displaystyle +\dots$

$\displaystyle +(1+b_n\epsilon+c_n\epsilon^2+\dots)\epsilon^{\alpha_n}A_nx^n$

Collecting terms according to their powers of $\epsilon$ , we obtain:

$\displaystyle P_{\epsilon}(x)=1+\epsilon^{\alpha_1}A_1x+\dots+\epsilon^{\alpha_n}A_nx^n$

$\displaystyle +b_0\epsilon+b_1\epsilon^{\alpha_1+1}A_1x+\dots+b_n\epsilon^{\alpha_n+1}A_nx^n$

$\displaystyle +\dots$

Theorem (Newton–Puiseux). Each root of the polynomial above is of the form:

$\displaystyle x(\epsilon)=\epsilon^pw(\epsilon),\quad w(0)\neq0$

where $w(\epsilon)$ is a continuous function near zero. For convenience, we simply write $x=\epsilon^pw$ . If the theorem holds, the polynomial above becomes:

$\displaystyle P_{\epsilon}(\epsilon^pw)=1+\epsilon^{\alpha_1}A_1\epsilon^pw+\dots+\epsilon^{\alpha_n}A_n(\epsilon^pw)^n$

$\displaystyle +b_0\epsilon+b_1\epsilon^{\alpha_1+1}A_1\epsilon^pw+\dots+b_n\epsilon^{\alpha_n+1}A_n(\epsilon^pw)^n$

$\displaystyle +\dots$

We define $Q_{\epsilon}(w):=1+\epsilon^{\alpha_1+p}A_1w+\dots+\epsilon^{\alpha_n+np}A_nw^n$ and have:

$\displaystyle P_{\epsilon}(\epsilon^pw)=Q_{\epsilon}(w)+\epsilon(b_0+b_1\epsilon^{\alpha_1+p}A_1w+\dots+b_n\epsilon^{\alpha_n+np}A_nw^n)+\dots$

If $\epsilon^pw$ is a root of $P_{\epsilon}(x)$ , we must satisfy:

$\displaystyle \lim_{\epsilon\to0}P_{\epsilon}(\epsilon^pw)=\lim_{\epsilon\to0}Q_{\epsilon}(w)=0$

Since $w(\epsilon)$ is a continuous function near zero, we inspect the structure of $Q_{\epsilon}(w(0))$ to determine the leading-order behavior:

$\displaystyle Q_{\epsilon}(w(0))=1+\epsilon^{\alpha_1+p}A_1w(0)+\dots+\epsilon^{\alpha_n+np}A_nw(0)^n$

Considering the set of its exponents of $\epsilon$ :

$\displaystyle E=\{0,\alpha_1+p,\dots,\alpha_k+kp,\dots,\alpha_n+np\}$

In order for the condition $\lim_{\epsilon\to0}Q_{\epsilon}(w)=0$ to hold, the minimal value of the set $E$ must be exactly $0$ , and this minimum must be shared by at least two distinct exponents. These identical minimal exponents define the dominant terms of $Q_{\epsilon}(w(0))$ which can then compensate each other to balance the equation at leading order.

July 2, 2026
Regularize II

$\displaystyle$

The underlying idea of regularization is to recover the roots that are lost in a singular limit. When some coefficients of a polynomial are multiplied by powers of a small parameter $\epsilon$ , the limiting polynomial obtained by setting $\epsilon=0$ may have fewer roots than the original polynomial. In this situation, some roots have escaped to a different scale and are therefore invisible in the limit.

To recover these missing roots, we introduce a rescaling

$\displaystyle x:=\epsilon^p w$

where the exponent $p$ is chosen so that at least two dominant terms of the polynomial balance each other. Geometrically, this choice is determined from the Newton polygon (or Newton diagram) by examining the exponents of $\epsilon$ . After an additional normalization by a factor $\epsilon^{-m_j}$ , one obtains a new limiting polynomial in the variable $w$ . This reduced polynomial captures the asymptotic behavior of the roots on the corresponding scale and typically restores the correct number of roots.

In this sense, the singularity is not an intrinsic property of the problem but rather the consequence of observing the problem at an inappropriate scale.

As a simple example, consider

$\displaystyle P(x) = x^2 -x+1$

$\displaystyle P_\epsilon(x) =\epsilon x^2-x+1$

Setting $\epsilon=0$ gives

$\displaystyle P_0(x)=-x+1$

which has only one root, although the original polynomial is quadratic and therefore has two roots. To recover the missing root, let

$\displaystyle x:=\epsilon^{-1}w$

Substituting into the polynomial gives

$\displaystyle P_\epsilon(\epsilon^{-1}w) = \epsilon(\epsilon^{-2}w^2)-\epsilon^{-1}w+1 = \epsilon^{-1}(w^2-w+\epsilon)$

Multiplying by $\epsilon$ yields

$\displaystyle w^2-w+\epsilon$

Taking the limit $\epsilon\to0$ gives the regular polynomial

$\displaystyle w^2-w=w(w-1)$

which has two roots. One corresponds to the finite root seen in the original limit, while the second reveals a root located on the larger scale $x=O(\epsilon^{-1})$ . The rescaling has therefore recovered the root that was hidden in the singular limit.

June 25, 2026
Regularize I

We will now present (without justification) a method for regularizing a singular problem involving a polynomial. We want to find the roots of :

$\displaystyle P(x) = x^6 - x^4 - x^3 + 8$

We decide to perturb the above polynomial as follows:

$\displaystyle P_{\epsilon}(x) = \epsilon^2 x^6 - \epsilon x^4 - x^3 + 8$

For $\epsilon = 0$ we have:

$\displaystyle P_{\epsilon= 0}(x) = -x^3 + 8$

This polynomial has three roots. We have a singular problem since the original polynomial has six roots.

We can try to fix the problem using a change of variable:

$\displaystyle x := \epsilon^p w$

$\displaystyle P_{\epsilon}(\epsilon^p w) = \epsilon^2 (\epsilon^p w)^6 - \epsilon (\epsilon^p w)^4 - (\epsilon^p w)^3 + 8$

$\displaystyle = \epsilon^2 (\epsilon^{6p} w^6) - \epsilon (\epsilon^{4p} w^4) - (\epsilon^{3p} w^3) + 8$

$\displaystyle = \epsilon^{6p+2} w^6 - \epsilon^{4p+1} w^4 - \epsilon^{3p} w^3 + 8$

Now we collect the exponents of $\epsilon$ : ( $6p+2, 4p+1$ , $3p, 0$ ) and plot them in a graph as a function of p (see figure below). If we think of the lines on the graph as delimiting a figure in the plane, we can select the vertex of this figure with the largest p value and the smallest f(p) value. In our case, the point corresponding to this criterion is the point of intersection of the lines $6p+2$ and $3p$ . We thus obtain the point of intersection with coordinates ( $-2/3, -2$ ).

We will call the coordinates of the intersection point $p_j$ and $m_j$ respectively.

Now calculate:

$\displaystyle \epsilon^{-m_j} P_{\epsilon}(\epsilon^{p_j} w) = \epsilon^{2} P_{\epsilon}(\epsilon^{-\frac{2}{3}} w)$

$\displaystyle = \epsilon^2 (\epsilon^{6(-\frac{2}{3})+2} w^6 - \epsilon^{4(-\frac{2}{3})+1} w^4 - \epsilon^{3(-\frac{2}{3})} w^3 + 8)$

$\displaystyle = \epsilon^2 (\epsilon^{-2} w^6 - \epsilon^{-\frac{5}{3}} w^4 - \epsilon^{-2} w^3 + 8)$

$\displaystyle = w^6 - \epsilon^{\frac{1}{3}} w^4 - w^3 + 8 \epsilon^2$

This is now a regular problem since the new polynomial:

$\displaystyle P(w) = w^6 - w^3$

has 6 roots.

Using the procedure above we have regularized a singular problem involving a polynomial using the following sequence:

$\displaystyle P(x) \to P_{\epsilon}(x) \to P_{\epsilon}(\epsilon^p w) \to \epsilon^{-m_j} P_{\epsilon}(\epsilon^{p_j} w)$

June 18, 2026
A quintic equation
So far, we’ve introduced the perturbation technique, described regular and irregular problems and presented solutions to simple problems that we could have solved exactly without using perturbative series. We will now consider a problem that cannot be solved exactly. In this way, we hope to illustrate the power of this method.

We want to find a real zero of the following quintic equation:

$\displaystyle x^5 + x = 1$

We will proceed as usual (see previous sections):

$\displaystyle x^5 + \epsilon x = 1$

$\displaystyle (a_0 + a_1\epsilon + a_2 \epsilon^2 + \cdots)^5 + \epsilon (a_0 + a_1\epsilon + a_2 \epsilon^2 + \cdots) = 1$

Solving the unperturbed problem (where $\epsilon = 0$ ) we see that $a_0 = 1$ ( $a_0$ is always the answer to the unperturbed problem). We can use Python to expand the perturbative series and compute the coefficients (see code below). In this example we will calculate up to six coefficients.
```
from sympy import *

# set variables
a1 = Symbol('a1')
a2 = Symbol('a2')
a3 = Symbol('a3')
a4 = Symbol('a4')
a5 = Symbol('a5')
x = Symbol('x') # epsilon

# from the unperturbed problem we know that a0 = 1

# expand the quintic equation
equ = expand ((1 + a1*x + a2*x**2 + a3*x**3 + a4*x**4 + a5*x**5)** 5 + x*(1 + a1*x + a2*x**2 + a3*x**3 + a4*x**4 + a5*x**5))

# get the coefficients separately
equ.coeff(x)

5*a1+1

equ.coeff(x**2)

10*a1**2+a1+5*a2

# solve the system of equations to calculate coefficients
solution = solve([equ.coeff(x),equ.coeff(x**2),equ.coeff(x**3)], (a1, a2, a3))
solution

[(-1/5, -1/25, -1/125)]

solution = solve([equ.coeff(x),equ.coeff(x**2),equ.coeff(x**3),equ.coeff(x**4), equ.coeff(x**5)], (a1, a2, a3, a4, a5))
solution

[(-1/5, -1/25, -1/125, 0, 21/15625)])
```
Therefore (using the results above calculated using Python):

$\displaystyle a_0 = 1$

$\displaystyle a_1 = -\frac{1}{5}$

$\displaystyle a_2 = -\frac{1}{25}$

$\displaystyle a_3 = -\frac{1}{125}$

$\displaystyle a_4 = 0$

$\displaystyle a_5 = \frac{21}{15625}$

We have:

$\displaystyle x(\epsilon) = 1 - \frac{1}{5}\epsilon - \frac{1}{25}\epsilon^2 - \frac{1}{125}\epsilon^3 + 0\cdot\epsilon^4 + \frac{21}{15625}\epsilon^5 + \cdots$

An approximation (fifth-order approximation) of the real solution of $x^5 + x = 1$ is (setting $\epsilon = 1$ ):

$\displaystyle 1 - \frac{1}{5} - \frac{1}{25} - \frac{1}{125} + 0 + \frac{21}{15625} \approx 0.7533$
June 11, 2026
Illustration of a singular problem

Illustration of a singular problem:

$\displaystyle x^2-2x +1 = 0 \quad \text{Problem}$

$\displaystyle \epsilon x^2-2x +1 = 0 \quad \text{Insert }\epsilon$

The solutions to the perturbed problem are (see figure below):

$\displaystyle x_1(\epsilon) = \frac{1+ \sqrt{1-\epsilon}}{\epsilon}$

$\displaystyle x_2(\epsilon) = \frac{1- \sqrt{1-\epsilon}}{\epsilon}$

While one of the roots approaches $\frac{1}{2}$ as $\epsilon \to 0$ , the other goes to infinity. This is a manifestation of a singular behaviour. The problems above illustrate the importance of setting $\epsilon$ to a right position.

A solution of the regular problem $x^2-2x +\epsilon = 0$ is $x_1(\epsilon) = 1 - \sqrt{1- \epsilon}$ . We would like to calculate the corresponding perturbative series.
Using:

$\displaystyle (a+b)^n = \sum_{k=0}^{\infty} {n\choose k} a^{n-k}b^{k}$

$\displaystyle = a^n + n a^{n-1}b + \frac{n(n-1)}{2} a^{n-2}b^{2}+...+ \frac{n(n-1)\cdots(n-k+1)}{k!} a^{n-k}b^k +...$

We would like to find the perturbative series corresponding to:

$\displaystyle \sqrt{1-\epsilon} = (1-\epsilon)^{1/2}$

with $a=1$ , $b = -\epsilon$ , $n = \frac{1}{2}$ therefore:

$\displaystyle \sqrt{1-\epsilon} = 1 - \frac{1}{2}\epsilon - \frac{1}{8}\epsilon^2 - ...$

$\displaystyle 1- \sqrt{1-\epsilon} = \frac{1}{2}\epsilon + \frac{1}{8}\epsilon^2 + ...$

Here are the first series and their corresponding graphs:

$\displaystyle s_2 = \frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2$

$\displaystyle s_3 =\frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2 + \frac{1}{16} \epsilon^3$

$\displaystyle s_4 = \frac{1}{2} \epsilon + \frac{1}{8} \epsilon^2 + \frac{1}{16} \epsilon^3 + \frac{5}{128} \epsilon^4$

June 4, 2026
Regular Problem II

Illustration of another regular problem:

$\displaystyle x^2-2x -1 = 0 \quad \text{Problem}$

$\displaystyle x^2-2\epsilon x-1=0 \quad \text{Insert }\epsilon$

The solutions to the perturbed problem are (see figure below):

$\displaystyle x_1(\epsilon) = \frac{2\epsilon + \sqrt{4 \epsilon^2 + 4}}{2} = \epsilon + \sqrt{1 + \epsilon^2}$

$\displaystyle x_2(\epsilon) = \frac{2\epsilon - \sqrt{4 \epsilon^2 + 4}}{2} = \epsilon - \sqrt{1 + \epsilon^2}$

The figure above shows solutions of the problem $x^2-2\epsilon x-1 = 0$ using perturbation theory. For $\epsilon =0$ the solutions are -1 and 1 and for $\epsilon =1$ the solutions are exactly $1+\sqrt{2}$ and $1-\sqrt{2}$ .

$\displaystyle x_1(\epsilon) = 1 + \epsilon + \frac{1}{2}\epsilon^2 - \frac{1}{8}\epsilon^4 + \frac{1}{16}\epsilon^6 - \frac{5}{128}\epsilon^8 + O(\epsilon^{10})$

$\displaystyle x_2(\epsilon) = -1 + \epsilon - \frac{1}{2}\epsilon^2 + \frac{1}{8}\epsilon^4 - \frac{1}{16}\epsilon^6 + \frac{5}{128}\epsilon^8 + O(\epsilon^{10})$

Setting $\epsilon = 1$ :

$\displaystyle x_1(1) \approx 2 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} - \frac{5}{128} = 2.3984375$

$\displaystyle x_2(1) \approx 0 - \frac{1}{2} + \frac{1}{8} - \frac{1}{16} + \frac{5}{128} = -0.3984375$

This simple example clearly illustrates the application of regular perturbation theory to a quadratic equation, showing how the solutions vary continuously with the perturbation parameter $\epsilon$ .

May 28, 2026
Regular Problem I
In perturbation theory we distinguish regular and singular problems. In this section we aim to solve the original regular problem:

$\displaystyle x^2 - 2 = 0 \quad \text{Original problem}$

$\displaystyle x^2 - (1 + \epsilon) = 0 \quad \text{Insert epsilon}$

The solutions are:

$\displaystyle x_1(\epsilon) = -\sqrt{1+\epsilon}$

$\displaystyle x_2(\epsilon) = \sqrt{1+\epsilon}$

At $\epsilon=0$ the problem is solvable analytically:

$\displaystyle x_1 = -1, \quad x_2 = 1$

At $\epsilon=1$ we recover the solution of the original problem:

$\displaystyle x_1 = -\sqrt{2}, \quad x_2 = \sqrt{2}$

Figure. Regular problem: $x^2-(1+\epsilon)=0$ connecting $x^2-1=0$ to $x^2-2=0$ .

Characteristics of a regular perturbation problem:
- The solution depends smoothly on ε.
- The deformation from ε=0 to ε=1 is continuous.
- The problem at ε=0 is simple and solvable analytically.
- The original problem is recovered exactly at ε=1.
- No singularities or discontinuities appear.
- The solution admits a regular power-series expansion in ε.
We treat $\epsilon$ as a parameter to track the perturbation. The solution $x_2(\epsilon) = \sqrt{1+\epsilon}$ can be expressed as a Taylor series around the simple problem ( $\epsilon = 0$ ):

$\displaystyle x_2(\epsilon) = 1 + \frac{1}{2}\epsilon - \frac{1}{8}\epsilon^2 + \frac{1}{16}\epsilon^3 - \frac{5}{128}\epsilon^4 + \dots$

The original problem $x^2 - 2 = 0$ is recovered by evaluating this series at $\epsilon = 1$ . The smooth dependence on the parameter allows us to reach the target solution. For $\epsilon=1$ , the first few terms give:

$\displaystyle x_2(1) \approx 1 + 0.5 - 0.125 + 0.0625 - 0.039 = 1.3985$

which is already a good approximation $\sqrt{2} \approx 1.4142$ .

This example illustrates a regular homotopy connecting the simple problem $x^2 - 1 = 0$ to the target problem $x^2 - 2 = 0$ .
May 21, 2026

Polynomial expansion

Expanding series and sorting coefficients by hand can be tedious. The Python library sympy allows us to automate this process efficiently.

Polynomial expansion and coefficient extraction

from sympy import symbols, expand, collect

# define variables
a0, a1, a2, a3, x = symbols('a0 a1 a2 a3 x')

# expand the expression
expr = expand((a0 + a1*x + a2*x**2 + a3*x**3)**2)

print("Expanded expression:")
print(expr)

print("\nCollected by powers of x:")
print(collect(expr, x))

Output:

Expanded expression:
a0**2 + 2*a0*a1*x + 2*a0*a2*x**2 + 2*a0*a3*x**3
+ a1**2*x**2 + 2*a1*a2*x**3 + 2*a1*a3*x**4
+ a2**2*x**4 + 2*a2*a3*x**5 + a3**2*x**6

Collected by powers of x:
a0**2 + 2*a0*a1*x
+ (2*a0*a2 + a1**2)*x**2
+ (2*a0*a3 + 2*a1*a2)*x**3
+ (2*a1*a3 + a2**2)*x**4
+ 2*a2*a3*x**5
+ a3**2*x**6

We can also extract coefficients order by order:

coeffs = [expr.coeff(x, i) for i in range(7)]

for i, c in enumerate(coeffs):
    print(f"Coefficient of x^{i}:", c)

Output:

Coefficient of x^0: a0**2
Coefficient of x^1: 2*a0*a1
Coefficient of x^2: 2*a0*a2 + a1**2
Coefficient of x^3: 2*a0*a3 + 2*a1*a2
Coefficient of x^4: 2*a1*a3 + a2**2
Coefficient of x^5: 2*a2*a3
Coefficient of x^6: a3**2

This approach allows us to systematically extract and match coefficients order by order, which is particularly useful in perturbation theory.

May 14, 2026

sqrt(2) bis

Let us revisit the problem of computing the value of $\sqrt{2}$ , but this time we slightly modify our perturbative approach to improve convergence.

Previously, we expanded around $x=1$ , which is relatively far from the true value. Instead, we now choose a better starting point.

$\displaystyle x = \sqrt{2} \quad \text{Problem}$

We observe that:

$\displaystyle \left(\frac{3}{2}\right)^2 = \frac{9}{4} = 2.25 \quad \text{so it is close to } 2$

This suggests writing:

$\displaystyle x = \frac{3}{2} + \delta$

The correction $\delta$ is expected to be small, which improves the convergence of the perturbative expansion.

We now insert this into the equation:

$\displaystyle x^2 = 2$

$\displaystyle \left(\frac{3}{2} + \delta \right)^2 = 2$

$\displaystyle \frac{9}{4} + 3\delta + \delta^2 = 2$

Rearranging, we obtain:

$\displaystyle 3\delta + \delta^2 = -\frac{1}{4}$

We now introduce a perturbation parameter $\epsilon$ :

$\displaystyle 3\delta + \delta^2 = -\frac{1}{4}\varepsilon$

We seek a solution of the form:

$\displaystyle \delta = b_1 \varepsilon + b_2 \varepsilon^2 + b_3 \varepsilon^3 + \cdots$

Substituting into the equation and expanding:

$\displaystyle 3b_1\varepsilon + 3b_2\varepsilon^2 + b_1^2\varepsilon^2 + \cdots = -\frac{1}{4}\varepsilon$

Matching powers of $\varepsilon$ :

$\displaystyle \varepsilon^1: \quad 3b_1 = -\frac{1}{4} \Rightarrow b_1 = -\frac{1}{12}$

$\displaystyle \varepsilon^2: \quad 3b_2 + b_1^2 = 0 \Rightarrow b_2 = -\frac{1}{432}$

Thus:

$\displaystyle \delta = -\frac{1}{12}\varepsilon - \frac{1}{432}\varepsilon^2 + \cdots$

We finally obtain:

$\displaystyle x = \frac{3}{2} - \frac{1}{12}\varepsilon - \frac{1}{432}\varepsilon^2 + \cdots$

Setting $\varepsilon = 1$ :

$\displaystyle x \approx \frac{3}{2} - \frac{1}{12} - \frac{1}{432} = 1.41435$

which is already a very good approximation of $\sqrt{2} \approx 1.41421$ .

This example illustrates a key idea in perturbation theory:

The choice of the unperturbed problem strongly affects the convergence of the series.

By expanding around a value closer to the true solution, we obtain a much more efficient approximation with fewer terms.

May 7, 2026
sqrt(2)

Let’s imagine we don’t have a calculating machine or computer and we want to calculate the value of the root of 2 using perturbation theory. Remember that we always insert epsilon so that the problem is exactly solvable for epsilon equals zero. Here’s how we might proceed:

$\displaystyle x = \sqrt{2} \quad \text{Problem}$

$\displaystyle x^2 = 2$

$\displaystyle x^2 - 1 = 1$

$\displaystyle x^2 - \epsilon = 1 \quad \text{Insert }\epsilon$

$\displaystyle (a_0 + a_1 \epsilon + a_2 \epsilon^2 + ...)^2 - \epsilon = 1 \quad \text{Insert }x_{\epsilon}$

$\displaystyle a_0^2 + 2a_0 a_1 \epsilon + 2 a_0 a_2 \epsilon^2 + a_1^2 \epsilon^2 + ... - \epsilon = 1$

$\displaystyle a_0^2 + (2a_0 a_1 - 1) \epsilon + (2 a_0 a_2 + a_1^2) \epsilon^2 + ... = 1$

In this case, we keep only the powers of epsilon up to order two. We want to perform a perturbative calculation of order two.

$\displaystyle \implies a_0 = 1, a_1 = \frac{1}{2}, a_2 = -\frac{1}{8}, ...$

$\displaystyle x_{\epsilon} = 1 + \frac{1}{2} \epsilon - \frac{1}{8}\epsilon^2 + ...$

$\displaystyle x_{\epsilon = 1} = 1 + \frac{1}{2} - \frac{1}{8} + ...$

Let us solve the same problem but this time we keep the powers of epsilon up to order three.

$\displaystyle x = \sqrt{2} \quad \text{Problem}$

$\displaystyle x^2 = 2$

$\displaystyle x^2 - 1 = 1$

$\displaystyle x^2 - \epsilon = 1 \quad \text{Insert }\epsilon$

$\displaystyle (a_0 + a_1 \epsilon + a_2 \epsilon^2 + a_3 \epsilon^3 + ...)^2 - \epsilon = 1 \quad \text{Insert }x_{\epsilon}$

$\displaystyle a_0^2+a_1^2\epsilon^2+2a_0a_1\epsilon+2a_0a_2\epsilon^2+2a_0a_3\epsilon^3+2a_1a_2\epsilon^3 + ... - \epsilon = 1$

$\displaystyle a_0^2+(2a_0a_1 -1)\epsilon+(2a_0a_2 + a_1^2)\epsilon^2+(2a_0a_3+2a_1a_2)\epsilon^3 + ... = 1$

$\displaystyle \implies a_0 = 1, a_1 = \frac{1}{2}, a_2 = -\frac{1}{8}, a_3 = \frac{1}{16}$

$\displaystyle x_{\epsilon} = 1 + \frac{1}{2} \epsilon - \frac{1}{8}\epsilon^2 + \frac{1}{16}\epsilon^3 + ...$

$\displaystyle x_{\epsilon = 1} = 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16}$

This suggests the following (using the double factorial notation):

$\displaystyle \sqrt{2} = 1 + \sum_{k=1}^\infty (-1)^{k+1} \frac{(2k-3)!!}{(2k)!!}$

$\displaystyle = 1 + \frac{1}{2} - \frac{1}{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} - \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \cdots$

$\displaystyle = 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} - \frac{5}{128} + \frac{7}{256} + \cdots$

We could use Euler’s summation to speed up the convergence of this series. We observe that perturbing the initial problem in this form:

$\displaystyle x^2 - \epsilon = 1$

is similar to finding the roots of the polynomial

$\displaystyle x^2 - \epsilon -1 = 0$

$\displaystyle x^2 - (\epsilon +1) = 0$

whose solution is (see figure below)

$\displaystyle x= \sqrt{\epsilon+1}$

April 30, 2026